Citation for this paper:
Ahmad, B., Alghanmi, M., Alsaedi, A., Srivastava, H.M. & Ntouyas, S.K. (2019). The
Langevin Equation in Terms of Generalized Liouville–Caputo Derivatives with
Nonlocal Boundary Conditions Involving a Generalized Fractional Integral.
Mathematics, 7(6), 533.
http://dx.doi.org/10.3390/math7060533
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The Langevin Equation in Terms of Generalized Liouville–Caputo Derivatives with
Nonlocal Boundary Conditions Involving a Generalized Fractional Integral
Bashir Ahmad, Madeaha Alghanmi, Ahmed Alsaedi and Hari M. Srivastava and
Sotiris K. Ntouyas
June 2019
© 2019 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open
access article distributed under the terms and conditions of the Creative Commons
Attribution (CC BY) license (
http://creativecommons.org/licenses/by/4.0/
).
This article was originally published at:
http://dx.doi.org/10.3390/math7060533
mathematics
Article
The Langevin Equation in Terms of Generalized
Liouville–Caputo Derivatives with Nonlocal
Boundary Conditions Involving a Generalized
Fractional Integral
Bashir Ahmad1 , Madeaha Alghanmi1, Ahmed Alsaedi1and Hari M. Srivastava2,3,* and Sotiris K. Ntouyas1,4
1 Nonlinear Analysis and Applied Mathematics (NAAM)-Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia;
bashirahmad−qau@yahoo.com (B.A.); madeaha@hotmail.com (M.A.); aalsaedi@hotmail.com (A.A.); sntouyas@uoi.gr (S.K.N.)
2 Department of Mathematics and Statistics, University of Victoria, Victoria, BC V8W 3R4, Canada 3 Department of Medical Research, China Medical University Hospital, China Medical University,
Taichung 40402, Taiwan
4 Department of Mathematics, University of Ioannina, 451 10 Ioannina, Greece
* Correspondence: harimsri@math.uvic.ca
Received: 25 April 2019; Accepted: 6 June 2019; Published: 11 June 2019
Abstract:In this paper, we establish sufficient conditions for the existence of solutions for a nonlinear Langevin equation based on Liouville-Caputo-type generalized fractional differential operators of different orders, supplemented with nonlocal boundary conditions involving a generalized integral operator. The modern techniques of functional analysis are employed to obtain the desired results. The paper concludes with illustrative examples.
Keywords: Langevin equation; generalized fractional integral; generalized Liouville–Caputo derivative; nonlocal boundary conditions; existence; fixed point
1. Introduction
The topic of fractional calculus has emerged as an interesting area of investigation in view of its widespread applications in social sciences, engineering and technical sciences. Mathematical models based on fractional order differential and integral operators are considered to be more realistic and practical than their integer-order counterparts as such models can reveal the history of the ongoing phenomena in systems and processes. This branch of mathematical analysis is now very developed and covers a wide range of interesting results, for instance [1–7].
The Langevin equation is an effective tool of mathematical physics, which can describe processes like anomalous diffusion in a descent manner. Examples of such processes include price index fluctuations [8], harmonic oscillators [9], etc. A generic Langevin equation for noise sources with correlations also plays a central role in the theory of critical dynamics [10]. The nature of the quantum noise can be understood better by means of a generalized Langevin equation [11]. The role of the Langevin equation in fractional systems, such as fractional reaction-diffusion systems [12,13], is very rich and beautiful. The fractional analogue (also known as the stochastic differential equation) of the usual Langevin equation is suggested for systems in which the separation between microscopic and macroscopic time scales is not observed; for example, see [8]. In [14], the author investigated moments, variances, position and velocity correlation for a Riemann-Liouville-type fractional Langevin equation
in time and compared the results obtained with the ones derived for the same generalized Langevin equation involving the Liouville-Caputo fractional derivative. Some recent results on the Langevin equation with different boundary conditions can be found in the papers [15–20] and the references cited therein.
Motivated by the aforementioned work on the Langevin equation and its variants, in this paper, we introduce and study a new form of Langevin equation involving generalized Liouville-Caputo derivatives of different orders and solve it with nonlocal generalized fractional integral boundary conditions. In precise terms, we investigate the problem:
ρ cDa+α (ρcDa+β +λ)x(t) = f(t, x(t)), t∈J := [a, T], λ∈ R, x(a) =0, x(η) =0, x(T) =µρIγa+x(ξ), a<η<ξ<T, (1)
whereρcDαa+,ρcDβa+denote the Liouville–Caputo-type generalized fractional differential operators of
order 1<α≤2, 0<β<1, ρ>0, respectively,ρIa+γ is the generalized fractional integral operator of
order γ>0 and ρ>0, and f :[a, T] × R → Ris a given continuous function.
Here, we emphasize that the present work may have useful applications in fractional quantum mechanics and fractional statistical mechanics, in relation to further generalization of the Feynman and Weiner path integrals [21].
We compose the rest of the article as follows. Section2contains the basic concepts of generalized fractional calculus and an auxiliary lemma dealing with the linear variant of the given problem. In Section3, we present the main results and illustrative examples.
2. Preliminaries
Definition 1([22]). The generalized left-sided fractional integral of order β>0 and ρ>0 of g∈Xcp(a, b)for
−∞<a<t<b<∞, is defined by: (ρIβ a+g)(t) = ρ1−β Γ(β) Z t a sρ−1 (tρ−sρ)1−βg(s)ds, (2)
where Xcp(a, b)denotes the space of all complex-valued Lebesgue measurable functions φ on(a, b)equipped with
the norm: kφkXp c = Z b a |x c φ(x)|pdx x 1/p <∞, c∈ R, 1≤ p≤∞.
Similarly, the right-sided fractional integralρIβ
b−g is defined by: (ρIβ b−g)(t) = ρ1−α Γ(β) Z b t sρ−1 (sρ−tρ)1−βg(s)ds. (3)
Definition 2([23]). For β> 0, n = [β] +1, ρ> 0 and 0 ≤ a< x < b < ∞, we define the generalized
fractional derivatives in terms of the generalized fractional integrals (2) and (3) as:
(ρDβ a+g)(t) = t1−ρ d dt n (ρIn−β a+ g)(t) = ρ β−n+1 Γ(n−β) t1−ρ d dt nZ t a sρ−1 (tρ−sρ)β−n+1g(s)ds, (4) and: (ρDβ b−g)(t) = −t1−ρd dt n (ρIn−β b− g)(t)
Mathematics 2019, 7, 533 3 of 10 = ρ β−n+1 Γ(n−β) −t1−ρd dt nZ b t sρ−1 (sρ−tρ)β−n+1g(s)ds, (5)
if the integrals in the above expressions exist.
Definition 3([24]). For β > 0, n = [β] +1 and g ∈ ACδn[a, b], the Liouville–Caputo-type generalized
fractional derivativesρcDβa+g and
ρ
cDb−β g are respectively defined via (4) and (5) as follows:
ρ cDa+β g(x) =ρDβa+ h g(t) − n−1
∑
k=0 δkg(a) k! tρ−aρ ρ ki (x), δ=x1−ρ d dx, (6) ρ cDb−β g(x) =ρDb−β h g(t) − n−1∑
k=0 (−1)kδkg(b) k! bρ−tρ ρ ki (x), δ=x1−ρ d dx, (7) where ACδn[a, b]denotes the class of all absolutely-continuous functions g possessing δn−1-derivative(δn−1g∈AC([a, b],R)), equipped with the normkgkACn
δ =∑ n−1
k=0kδkgkC.
Remark 1([24]). For α≥0 and g∈ ACnδ[a, b], the left and right generalized Liouville–Caputo derivatives of g are respectively defined by the expressions:
ρ cDa+β g(t) = 1 Γ(n−β) Z t a tρ−sρ ρ n−β−1(δng)(s)ds s1−ρ , (8) ρ cDβb−g(t) = 1 Γ(n−β) Z b t sρ−tρ ρ n−α−1(−1)n(δng)(s)ds s1−ρ . (9)
Lemma 1([24]). Let g∈ACδn[a, b]or Cδn[a, b]and β∈ R. Then:
ρIβ a+ ρ cDβa+g(x) =g(x) − n−1
∑
k=0 (δkg)(a) k! xρ−aρ ρ k , ρIβ b− ρ cDb−β g(x) =g(x) − n−1∑
k=0 (−1)k(δkg)(a) k! bρ−xρ ρ k .In particular, for 0<β≤1, we have:
ρIβ a+ ρ cDβa+g(x) =g(x) −g(a), ρI β b− ρ cDbβ−g(x) =g(x) −g(b).
Definition 4. A function x∈C([a, T],R)is called a solution of (1) if x satisfies the equationρcDαa+(
ρ
cDa+β +
λ)x(t) = f(t, x(t))on[a, T], and the conditions x(a) =0, x(η) =0, x(T) =µρIa+γ x(ξ).
In the next lemma, we solve the linear variant of Problem (1).
Lemma 2. Let h∈C([a, T],R), x∈ AC3δ(J)and:
Ω=h(T ρ−aρ)β(Tρ−ηρ) ρβ+1Γ(β+2) −µ(ξρ−aρ)β+γ[(β+1)(ξρ−ηρ) −γ(ηρ−aρ)] ρβ+γ+1Γ(β+γ+2)(β+1) i 6=0. (10)
Then, the unique solution of linear problem: ρ cDαa+(ρcDa+β +λ)x(t) =h(t), t∈ J := [a, T], x(a) =0, x(η) =0, x(T) =µρIa+γ x(ξ), a<η<ξ<T, (11)
is given by: x(t) = ρIα+β a+ h(t) −λρI β a+x(t) + (tρ−aρ)β(ηρ−tρ) ρβ+1Γ(β+2)Ω n ρIα+β a+ h(T) −λρI β a+x(T) −µρIa+α+β+γh(ξ) +µλρIa+β+γx(ξ) o − (tρ−aρ)β Ω(ηρ−aρ)β (Tρ−aρ)β(Tρ−tρ) ρβ+1Γ(β+2) −µ(ξρ−aρ)β+γ[(β+1)(ξρ−tρ) −γ(tρ−aρ)] ρβ+γ+1Γ(β+γ+2)(β+1) n ρIα+β a+ h(η) −λρI β a+x(η) o . (12) Proof. ApplyingρIα
a+on the fractional differential equation in (11) and using Lemma1yield:
(ρcDa+β +λ)x(t) =ρIa+α h(t) +c1+c2
(tρ−aρ)
ρ , (13)
for some c1, c2∈ R.
ApplyingρIβ
a+to both sides of Equation (13), the general solution of the Langevin equation in (11)
is found to be: x(t) =ρIα+β a+ h(t) −λρI β a+x(t) +c1 (tρ−aρ)β ρβΓ(β+1) +c2 (tρ−aρ)β+1 ρβ+1Γ(β+2) +c3, (14) where c3∈ R.
Using the condition x(a) =0 in (14), we find that c3=0. Inserting the value of c3in (14) and then
applying the operatorρIγ
a+on the resulting equation, we get:
ρIγ ax(t) =ρIa+α+β+γh(t) −λρI β+γ a+ x(t) +c1 (tρ−aρ)β+γ ρβ+γΓ(β+γ+1) +c2 (tρ−aρ)β+γ+1 ρβ+γ+1Γ(β+γ+2). (15)
Using the boundary conditions x(η) =0 and x(T) =µρIa+γ x(ξ)together with (14) and (15) leads
to a system of algebraic equations in c1and c2, which, upon solving, yields:
c1 = ρβΓ(β+1) Ω(ηρ−aρ)β n(ηρ−aρ)β+1 ρβ+1Γ(β+2) ρIα+β a+ h(T) −λρIa+β x(T) −µρIα+β+γh(ξ) +µλρIβ+γx(ξ) −(T ρ−aρ)β+1 ρβ+1Γ(β+2)− µ(ξρ−aρ)β+γ+1 ρβ+γ+1Γ(β+γ+2) ρIα+β a+ h(η) −λρIa+β x(η) o , c2 = − ρβΓ(β+1) Ω(ηρ−aρ)β n(ηρ−aρ)β ρβΓ(β+1) ρIα+β a+ h(T) −λρIa+β x(T) −µρIα+β+γh(ξ) +µλρIβ+γx(ξ) −(T ρ−aρ)β ρβΓ(β+1)− µ(ξρ−aρ)β+γ ρβ+γΓ(β+γ+1) ρIα+β a+ h(η) −λρI β a+x(η) o .
Inserting the values of c1, c2and c3in (13) yields the solution (12). The converse of the Lemma2,
can be obtained by direct computation. This finishes the proof.
3. Existence and Uniqueness Results
In view of Lemma2, we introduce an operatorF :C → Cby:
F (x)(t) = ρIα+β a+ f(t, x(t)) −λρIa+β x(t) + (tρ−aρ)β(ηρ−tρ) ρβ+1Γ(β+2)Ω n ρIα+β a+ f(T, x(T)) −λρIa+β x(T) − µρIa+α+β+γf(ξ, x(ξ)) +µλρIa+β+γx(ξ) o − (t ρ−aρ)β Ω(ηρ−aρ)β h(Tρ−aρ)β(Tρ−tρ) ρβ+1Γ(β+2) − µ(ξρ−aρ)β+γ[(β+1)(ξρ−tρ) −γ(tρ−aρ)] ρβ+γ+1Γ(β+γ+2)(β+1) in ρIα+β a+ f(η, x(η)) −λρI β a+x(η) o . (16)
Mathematics 2019, 7, 533 5 of 10
Here,Cdenotes the Banach space of all continuous functions from[a, T]toRequipped with the normkxk =supt∈[a,T]|x(t)|.
For the sake of computational convenience, we set:
Λ1 = (Tρ−aρ)α+β ρα+βΓ(α+β+1) h 1+ ζ1 ρβ+1Γ(β+2)|Ω| i + |µ|(ξ ρ−aρ)α+β+γζ 1 ρα+2β+γ+1Γ(α+β+γ+1)Γ(β+2)|Ω| + (η ρ−aρ)αζ 2 ρα+βΓ(α+β+1)|Ω|, (17) Λ2 = |λ|(Tρ−aρ)β ρβΓ(β+1) h 1+ ζ1 ρβ+1Γ(β+2)|Ω| i + |µ||λ|(ξ ρ−aρ)β+γζ 1 ρ2β+γ+1Γ(β+γ+1)Γ(β+2)|Ω| + |λ|ζ2 ρβΓ(β+1)|Ω|, (18) where: ζ1:= max t∈[a,T] (t ρ−aρ)β(ηρ−tρ) , (19) ζ2:= max t∈[a,T] (t ρ−aρ)βh(Tρ−aρ)β(Tρ−tρ) ρβ+1Γ(β+2) −µ(ξρ−aρ)β+γ[(β+1)(ξρ−tρ) −γ(tρ−aρ)] ρβ+γ+1Γ(β+γ+2)(β+1) i . (20)
Now, we are in a position to present our main results. Our first existence result for the problem (1) is based on Krasnoselskii’s fixed point theorem [25], which is stated below.
Lemma 3. (Krasnoselskii’s fixed point theorem) LetS be a closed convex and non-empty subset of a Banach space E. LetG1,G2be the operators fromS to E such that(a) G1x+ G2y ∈ Swhenever u, v∈ S;(b) G1is
compact and continuous; and(c) G2is a contraction mapping. Then, there exists a fixed point ω∈ Ssuch that
ω= G1ω+ G2ω.
Theorem 1. Let f : J× R → Rbe a continuous function such that the following condition holds: (A1) There exists a continuous function φ∈C([a, T],R+)such that:
|f(t, u)| ≤φ(t), ∀(t, u) ∈J× R.
Then, the problem (1) has at least one solution on J, provided that:
Λ2<1. (21)
Proof. Introduce a closed ball Br= {x ∈ C:kxk ≤r}, with r> kφkΛ1−Λ21,kφk =supt∈[a,T]|φ(t)|, where
Λ2is given by (18). Then, we define operatorsF1andF2from Br toCby:
F1(x)(t) = ρIα+β a+ f(t, x(t)) + (tρ−aρ)β(ηρ−tρ) ρβ+1Γ(β+2)Ω n ρIα+β a+ f(T, x(T)) −µρI α+β+γ a+ f(ξ, x(ξ)) o − (tρ−aρ)β Ω(ηρ−aρ)β h(Tρ−aρ)β(Tρ−tρ) ρβ+1Γ(β+2) −µ(ξρ−aρ)β+γ[(β+1)(ξρ−tρ) −γ(tρ−aρ)] ρβ+γ+1Γ(β+γ+2)(β+1) i ×ρIα+β a+ f(η, x(η)), F2(x)(t) = −λρIa+β x(t) − (t ρ−aρ)β(ηρ−tρ) ρβ+1Γ(β+2)Ω n λρIa+β x(T) −µλρIa+β+γx(ξ) o + λ(t ρ−aρ)β Ω(ηρ−aρ)β × ×h(Tρ−aρ)β(Tρ−tρ) ρβ+1Γ(β+2) −µ(ξρ−aρ)β+γ[(β+1)(ξρ−tρ) −γ(tρ−aρ)] ρβ+γ+1Γ(β+γ+2)(β+1) i ρIβ a+x(η).
Note thatF = F1+ F2on Br. For x, y∈Br, we find that: kF1x+ F2yk ≤ sup t∈J n ρIα+β a+ |f(t, x(t))| + |λ|ρI β a+|y(t)| +(t ρ−aρ)β|(ηρ−tρ)| ρβ+1Γ(β+2)|Ω| n ρIα+β a+ |f(T, x(T))| + |λ|ρI β a+|y(T)| +|µ|ρIα+β+γa+ |f(ξ, x(ξ))| + |µ||λ|ρIa+β+γ|x(ξ)| o + (t ρ−aρ)β |Ω|(ηρ−aρ)β (Tρ−aρ)β(Tρ−tρ) ρβ+1Γ(β+2) −µ(ξρ−aρ)β+γ[(β+1)(ξρ−tρ) −γ(tρ−aρ)] ρβ+γ+1Γ(β+γ+2)(β+1) n ρIα+β a+ |f(η, x(η))| + |λ|ρI β a+|y(η)| o ≤ kφk ( (Tρ−aρ)α+β ρα+βΓ(α+β+1) h 1+ ζ1 ρβ+1Γ(β+2)|Ω| i + |µ|(ξ ρ−aρ)α+β+γζ 1 ρα+2β+γ+1Γ(α+β+γ+1)Γ(β+2)|Ω| + (η ρ−aρ)αζ 2 ρα+βΓ(α+β+1)|Ω| ) + kxkn|λ|(Tρ−aρ)β ρβΓ(β+1) h 1+ ζ1 ρβ+1Γ(β+2)|Ω| i + |µ||λ|(ξ ρ−aρ)β+γζ 1 ρ2β+γ+1Γ(β+γ+1)Γ(β+2)|Ω|+ |λ|ζ2 ρβΓ(β+1)|Ω| ) ≤ kφkΛ1+rΛ2<r. Thus,F1x+ F2y∈Br.
Next, it will be shown thatF2is a contraction. For that, let x, y∈ C. Then:
kF2x− F2yk ≤ sup t∈J ( |λ|ρIa+β |x(t) −y(t)| +|(t ρ−aρ)β(ηρ−tρ)| ρβ+1Γ(β+2)|Ω| × ×n|λ|ρIa+β |x(T) −y(T)| + |µ||λ|ρIa+β+γ|x(ξ) −y(ξ)| o +|λ||(t ρ−aρ)β| |Ω|(ηρ−aρ)β (Tρ−aρ)β(Tρ−tρ) ρβ+1Γ(β+2) − µ(ξρ−aρ)β+γ[(β+1)(ξρ−tρ) −γ(tρ−aρ)] ρβ+γ+1Γ(β+γ+2)(β+1) × ×ρIβ a+|x(η) −y(η)| ) ≤ n|λ|(Tρ−aρ)β ρβΓ(β+1) h 1+ ζ1 ρβ+1Γ(β+2)|Ω| i + |µ||λ|(ξ ρ−aρ)β+γζ 1 ρ2β+γ+1Γ(β+γ+1)Γ(β+2)|Ω| + |λ|ζ2 ρβΓ(β+1)|Ω| o kx−yk = Λ2kx−yk,
which, by the condition (21), implies thatF2is a contraction. The continuity of the operatorF1follows
from that of f . Furthermore,F1is uniformly bounded on Bras:
kF1xk ≤ kφkΛ1.
Finally, we establish the compactness of the operatorF1. Let us set sup(t,x)∈J×Br|f(t, x)| = ¯f<∞.
Then, for t1, t2∈ J, t1<t2, we have:
Mathematics 2019, 7, 533 7 of 10 = ρ1−(α+β) Γ(α+β) hZ t1 0 s ρ−1[(tρ 2−sρ)α+β−1− (t ρ 1−sρ)α+β−1]f(s, x(s))ds + Z t2 t1 sρ−1(tρ 2−sρ)α+β−1f(s, x(s))ds i +h(t ρ 2−aρ)β(ηρ−t ρ 2) ρβ+1Γ(β+2)Ω − (tρ1−aρ)β(ηρ−tρ 1) ρβ+1Γ(β+2)Ω in ρIα+β a+ f(T, x(T)) −µρI α+β+γ a+ f(ξ, x(ξ)) o −h (t ρ 2−aρ)β Ω(ηρ−aρ)β h(Tρ−aρ)β(Tρ−tρ 2) ρβ+1Γ(β+2) − µ(ξρ−aρ)β+γ[(β+1)(ξρ−tρ2) −γ(tρ−aρ)] ρβ+γ+1Γ(β+γ+2)(β+1) i − (t ρ 1−aρ)β Ω(ηρ−aρ)β h(Tρ−aρ)β(Tρ−tρ 1) ρβ+1Γ(β+2) − µ(ξρ−aρ)β+γ[(β+1)(ξρ−tρ1) −γ(tρ1−aρ)] ρβ+γ+1Γ(β+γ+2)(β+1) ii × ×ρIα+β a+ f(η, x(η)) ≤ ¯f ρα+βΓ(α+β+1) n |tρ(α+β)2 −tρ(α+β)1 | +2(tρ2−tρ1)α+βo + (tρ2−aρ)β(ηρ−tρ 2) ρβ+1Γ(β+2)Ω −(t ρ 1−aρ)β(ηρ−t ρ 1) ρβ+1Γ(β+2)Ω n ρIα+β a+ |f(T, x(T))| +µρI α+β+γ a+ |f(ξ, x(ξ))| o + (tρ2−aρ)β Ω(ηρ−aρ)β h(Tρ−aρ)β(Tρ−tρ 2) ρβ+1Γ(β+2) −µ(ξ ρ−aρ)β+γ[(β+1)(ξρ−tρ 2) −γ(t ρ 2−aρ)] ρβ+γ+1Γ(β+γ+2)(β+1) i − (t ρ 1−aρ)β Ω(ηρ−aρ)β h(Tρ−aρ)β(Tρ−tρ 1) ρβ+1Γ(β+2) − µ(ξ ρ−aρ)β+γ[(β+1)(ξρ−tρ 1) −γ(t ρ 1−aρ)] ρβ+γ+1Γ(β+γ+2)(β+1) i × ×ρIα+β a+ |f(η, x(η))|,
which tends to zero as t2→t1, independently of x∈ Br. Thus,F1is equicontinuous. Therefore,F1is
relatively compact on Br. As a consequence, we deduce by the the Arzelá–Ascoli theorem thatF1is
compact on Br. Thus, the hypothesis of Lemma3is satisfied. Therefore, the conclusion of Lemma3
applies, and hence, there exists at least one solution for the problem (1) on J.
In the next result, the uniqueness of solutions for the problem (1) is shown by means of the Banach contraction mapping principle.
Theorem 2. Let f : J× R → Rbe a continuous function satisfying the Lipschitz condition: (A2)
|f(t, u) −f(t, v)| ≤L|u−v|, L>0, /, for t∈ J and every u, v∈ R. Then, there exists a unique solution for the problem (1) on[a, T], provided that:
LΛ1+Λ2<1, (22)
whereΛ1andΛ2are respectively given by (17) and (18).
Proof. In the first step, we show thatFB¯r ⊂ B¯r, where B¯r = {x ∈ C([a, T],R) : kxk ≤ ¯r}, M = supt∈[a,T]|f(t, 0)|, ¯r≥ Λ1M
1−LΛ1−Λ2, and the operatorF :C → Cis given by (16). For x∈ B¯r, using(A2),
we get: |F (x)(t)| ≤ ρIα+β a+ [|f(t, x(t)) −f(t, 0)| + |f(t, 0)|] + |λ|ρI β a+|x(t)| +(t ρ−aρ)β|(ηρ−tρ)| ρβ+1Γ(β+2)|Ω| n ρIα+β a+ [|f(T, x(T)) −f(T, 0)| + |f(T, 0)|] + |λ|ρIa+β |x(T)| +|µ|ρIa+α+β+γ[|f(ξ, x(ξ)) −f(ξ, 0)| + |f(ξ, 0)|] + |µ||λ|ρIa+β+γ|x(ξ)| o
+ (t ρ−aρ)β |Ω|(ηρ−aρ)β (Tρ−aρ)β(Tρ−tρ) ρβ+1Γ(β+2) −µ(ξ ρ−aρ)β+γ[( β+1)(ξρ−tρ) −γ(tρ−aρ)] ρβ+γ+1Γ(β+γ+2)(β+1) ×nρIα+β a+ [|f(η, x(η)) −f(η, 0)| + |f(η, 0)|] + |λ|ρI β a+|x(η)| o ≤ (L¯r+M) (T ρ−aρ)α+β ρα+βΓ(α+β+1) h 1+ ζ1 ρβ+1Γ(β+2)|Ω| i + |µ|(ξ ρ−aρ)α+β+γζ 1 ρα+2β+γ+1Γ(α+β+γ+1)Γ(β+2)|Ω| + (η ρ−aρ)αζ 2 ρα+βΓ(α+β+1)|Ω| +¯r|λ|(T ρ−aρ)β ρβΓ(β+1) h 1+ ζ1 ρβ+1Γ(β+2)|Ω| i + |µ||λ|(ξ ρ−aρ)β+γζ 1 ρ2β+γ+1Γ(β+γ+1)Γ(β+2)|Ω| + |λ|ζ2 ρβΓ(β+1)|Ω| = (L¯r+M)Λ1+Λ2¯r≤¯r,
which, on taking the norm for t ∈ [a, T], implies thatkF (x)k ≤ ¯r. Thus, the operatorF maps B¯r
into itself. Now, we proceed to prove that the operatorF is a contraction. For x, y∈C([a, T],R)and t∈ [a, T], we have: |F (x)(t) − F (y)(t)| ≤ ρIα+β a+ |f(t, x(t)) −f(t, y(t))| + |λ|ρI β a+|x(t) −y(t)| +(t ρ−aρ)β|(ηρ−tρ)| ρβ+1Γ(β+2)|Ω| n ρIα+β a+ |f(T, x(T)) −f(T, y(T))| + |λ|ρIa+β |x(T) −y(T)| +|µ|ρIa+α+β+γ|f(ξ, x(ξ)) −f(ξ, y(ξ))| + |µ||λ|ρIa+β+γ|x(ξ) −y(ξ)| o + (t ρ−aρ)β |Ω|(ηρ−aρ)β (Tρ−aρ)β(Tρ−tρ) ρβ+1Γ(β+2) − µ(ξρ−aρ)β+γ[(β+1)(ξρ−tρ) −γ(tρ−aρ)] ρβ+γ+1Γ(β+γ+2)(β+1) × ×nρIα+β a+ |f(η, x(η)) −f(η, y(η))| + |λ|ρIa+β |x(η) −y(η)| o ≤ Lkx−yk (T ρ−aρ)α+β ρα+βΓ(α+β+1) h 1+ ζ1 ρβ+1Γ(β+2)|Ω| i + |µ|(ξ ρ−aρ)α+β+γζ 1 ρα+2β+γ+1Γ(α+β+γ+1)Γ(β+2)|Ω| + (η ρ−aρ)αζ 2 ρα+βΓ(α+β+1)|Ω| + kx−yk|λ|(T ρ−aρ)β ρβΓ(β+1) h 1+ ζ1 ρβ+1Γ(β+2)|Ω| i + |µ||λ|(ξ ρ−aρ)β+γζ 1 ρ2β+γ+1Γ(β+γ+1)Γ(β+2)|Ω| + |λ|ζ2 ρβΓ(β+1)|Ω| = (LΛ1+Λ2)kx−yk.
Taking the norm of the above inequality for t∈ [a, T], we get: kF (x) − F (y)k ≤ (LΛ1+Λ2)kx−yk,
which implies that the operatorFis a contraction on account of the condition (22). Thus, we deduce by the Banach contraction mapping principle that the operatorF has a unique fixed point. Hence, there exists a unique solution for the problem (1). The proof is complete.
Example 1. Let us consider the following boundary value problem:
1/3 c D5/4 1/3 c D1/4+1/5 x(t) = √ 1 400+t |x(t)| +2 |x(t)| +1 +e −t , t∈ J := [1, 2], x(1) =0, x(3/2) =0, x(2) =2/71/3I3/4x(7/4). (23)
Mathematics 2019, 7, 533 9 of 10 Here, ρ = 1/3, α = 5/4, β = 1/4, γ = 3/4, λ = 1/5, µ = 2/7, a = 1, η = 3/2, ξ = 7/4, T = 2 and f(t, x) = √ 1 400+t |x|+2 |x|+1 +e−t
. Using the given data, we find that |Ω| ≈ 0.293634,Λ1 ≈
1.336009,Λ2≈0.673563, ζ1≈0.082260, ζ2≈0.232036, whereΩ, Λ1,Λ2, ζ1, and ζ2are given by (10),
(17), (18), (19) and (20) respectively.
For illustrating Theorem1, we show that all the conditions of Theorem1are satisfied. Clearly, f(t, x) is continuous and satisfies the condition (A1) with φ(t) = 2+e
−t
√
400+t. Furthermore, Λ2 ≈
0.673563<1. Thus, all the conditions of Theorem1are satisfied, and consequently, the problem (23) has at least one solution on[1, 2].
Furthermore, Theorem2 is applicable to the problem (23) with L = 1/20 as LΛ1+Λ2 ≈
0.740363 < 1. Thus, all the assumptions of Theorem2are satisfied. Therefore, the conclusion of Theorem2applies to the problem (23) on[1, 2].
4. Conclusions
We have introduced a new type of nonlinear Langevin equation in terms of Liouville-Caputo-type generalized fractional differential operators of different orders and solved it with nonlocal generalized integral boundary conditions. The existence result was obtained by applying the Krasnoselskii fixed point theorem without requiring the nonlinear function to be of the Lipschitz type, while the uniqueness of solutions for the given problem was based on a celebrated fixed point theorem due to Banach. Here, we remark that many known existence results, obtained by means of the Krasnoselskii fixed point theorem, demand the associated nonlinear function to satisfy the Lipschitz condition. Moreover, by fixing the parameters involved in the given problem, we can obtain some new results as special cases of the ones presented in this paper. For example, letting ρ=1, µ=0, a=0 and T=1 in the results of Section3, we get the ones derived in [15].
Author Contributions:Formal analysis, B.A., M.A., A.A., H.M.S. and S.K.N.
Funding:This project was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, Saudi Arabia, under Grant No. KEP-PhD-24-130-40. The authors acknowledge with thanks the DSR’s technical and financial support. The authors also acknowledge the reviewers for their constructive remarks on our work.
Conflicts of Interest:The authors declare no conflict of interest. References
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